URAL 2037 Richness of binary words

B - Richness of binary words

Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 2037

Description

For each integer i from 1 to n, you must print a string s_i of length n consisting of letters ‘a’ and ‘b’ only. The string s_i must contain exactly i distinct palindrome substrings. Two substrings are considered distinct if they are different as strings.

Input

The input contains one integer n (1 ≤ n ≤ 2000).

Output

You must print n lines. If for some i, the answer exists, print it in the form “ i : s_i” where s_i is one of possible strings. Otherwise, print “i : NO”.

Sample Input

input	output
4	1 : NO 2 : NO 3 : NO 4 : aaaa

一个字符串只能由字符a b组成给定一个数字n 问从1~n 能不能有一个字符串使得这个字符串里面的回文子串的个数有i个有就输出这个字符串否则就输出NO

一开始模拟了很多样例。。。对于长度为n的字符串只能找到回文子串个数为n的所以就猜想除了n 其他都是NO

然后就WA了

后来暴力了之后发现一个字符串出现的比较频繁 abbaba

后来发现这是一个循环结整个的字符串的个数是跟这段字符串有关的。。。

我是搬运工

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)using namespace std;inline int Read()
{char ch;int a = 0;while((ch = getchar()) == ' ' | ch == '\n');a += ch - '0';while((ch = getchar()) != ' ' && ch != '\n'){a *= 10;a += ch - '0';}return a;
}inline void Print(int a)
{if(a>9)Print(a/10);putchar(a%10+'0');
}char ss[] = "abbaba";int main()
{//fread;int n;while(~scanf("%d", &n)){if(n <= 7){for(int i = 1; i < n; i++){Print(i);puts(" : NO");}Print(n);putchar(' '), putchar(':'), putchar(' ');for(int i = 1; i <= n; i++) putchar('a');puts("");}else if(n == 8){for(int i = 1; i < 7; i++){Print(i);puts(" : NO");}puts("7 : aabbabaa");Print(n);putchar(' '), putchar(':'), putchar(' ');for(int i = 1; i <= n; i++) putchar('a');puts("");}else{for(int i = 1; i <= 7; i++){Print(i);puts(" : NO");}for(int i = 8; i < n; i++){Print(i);putchar(' '), putchar(':'), putchar(' ');for(int j = 0; j < i - 8 + 1; j++) putchar('a');int loop = (n - (i - 8 + 1)) / 6, res = (n - (i - 8 + 1)) % 6;while(loop--){
//                    printf("abbaba");putchar('a');putchar('b');putchar('b');putchar('a');putchar('b');putchar('a');}for(int j = 0; j < res; j++) putchar(ss[j]);puts("");}Print(n);putchar(' '), putchar(':'), putchar(' ');
//            printf("%d : ", n);for(int i = 1; i <= n; i++) putchar('a');puts("");}}return 0;
}

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